lg problem

This is a bit of chalange for me and that is why I am posting it here The problem says: Prove that lg(lgX) > 2500, where X equals (2^10000)! . The ! sign means factoriel. Please if you dont know how to solve it atleast write where you got after trying so we could solve this problem together. Thanks.

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Re: lg problem

The value of (2^10000)! can be found approximately using the James-Stirling formula,where n! ~ [√(2*pi*n)]*[(n/e)^n]2^10,000 contain 3011 digits.The factorial of this number would be of the order of 10^10^3011 (actually, much greater!).Log to base 10 of 10^10^3011 is 10^3011and log of log of 10^10^3011 is 3011, which is greater than 2500!q.e.d

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Re: lg problem

2^10,000 is approximate 10^3011.We know that for any number greater than 25, n!>10^n.Therefore, (2^10,000)! would be greater than 10^10^3,011.You can omit the √(2*pi*n) in the James Stirling formula as it wouldn't make a big difference.If you are only required to show lg(lgX) > 2500, where X equals (2^10000)!, the James Stirling formula may not be necessary, as the simpler method has been discussed. Don't be sorry for seeking clarifications to your doubts.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Re: lg problem

No, not a theorem. An observation that can be proved.By using the calculator (scientific mode), you can see that 25! > 10^25.Therefater, the LHS of the inequation increases by 26, 27, 28 etc. for 26!, 27!, 28!. On the other hand, the RHS increases only by 10 in each step. Therefore, it can be proved that for any number, n ≥ 25, n!>10^n.Wish you happy holidays too

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.